## Elementary Geometry for College Students (6th Edition)

Published by Brooks Cole

# Appendix A - A.4 - Quadratic Equations - Exercises - Page 550: 28

#### Answer

$x=\frac{3}{2}~~$ or $~~x=-2$

#### Work Step by Step

Let's consider a trinomial in this form: $~ax^2+bx+c$ To factor this trinomial, we need to find two numbers $r$ and $s$ such that $r+s = b$ and $r\times s = a\times c$ Then the next step is to write the trinomial as follows: $~x^2+bx+c = ax^2+rx+sx+c$ Let's consider this equation: $~2x^2+x-6 = 0$ To factor the left side of this equation, we need to find two numbers $r$ and $s$ such that $r+s = 1$ and $r\times s = -12$. We can see that $(4)+(-3) = 1~$ and $(4)\times (-3) = -12$ We can solve the equation as follows: $2x^2+x-6 = 0$ $2x^2+4x-3x-6 = 0$ $(2x^2+4x)+(-3x-6) = 0$ $(2x)(x+2)+(-3)(x+2) = 0$ $(2x-3)~(x+2) = 0$ $2x-3 = 0~~$ or $~~x+2 = 0$ $x=\frac{3}{2}~~$ or $~~x=-2$

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